I want to get only those rows that contain ONLY certain characters in a column.
Let's say the column name is DATA.
I want to get all rows where in DATA are ONLY (must have all three conditions!):
Numeric characters (1 2 3 4 5 6 7 8 9 0)
Dash (-)
Comma (,)
For instance:
Value "10,20,20-30,30" IS OK
Value "10,20A,20-30,30Z" IS NOT OK
Value "30" IS NOT OK
Value "AAAA" IS NOT OK
Value "30-" IS NOT OK
Value "30," IS NOT OK
Value "-," IS NOT OK
Try patindex:
select * from(
select '10,20,20-30,30' txt union
select '10,20,20-30,40' txt union
select '10,20A,20-30,30Z' txt
)x
where patindex('%[^0-9,-]%', txt)=0
For you table, try like:
select
DATA
from
YourTable
where
patindex('%[^0-9,-]%', DATA)=0
As per your new edited question, the query should be like:
select
DATA
from
YourTable
where
PATINDEX('%[^0-9,-]%', DATA)=0 and
PATINDEX('%[0-9]%', LEFT(DATA, 1))=1 and
PATINDEX('%[0-9]%', RIGHT(DATA, 1))=1 and
PATINDEX('%[,-][-,]%', DATA)=0
Edit: Your question was edited, so this answer is no longer correct. I won't bother updating it since someone else already has updated theirs. This answer does not fulfil the condition that all three character types must be found.
You can use a LIKE expression for this, although it's slightly convoluted:
where data not like '%[^0123456789,!-]%' escape '!'
Explanation:
[^...] matches any character that is not in the ... part. % matches any number (including zero) of any character. So [^0123456789-,] is the set of characters that you want to disallow.
However: - is a special character inside of [], so we must escape it, which we do by using an escape character, and I've chosen !.
So, you match rows that do not contain (not like) any character that is not in your disallowed set.
Use option with PATINDEX and LIKE logic operator
SELECT *
FROM dbo.test70
WHERE PATINDEX('%[A-Z]%', DATA) = 0
AND PATINDEX('%[0-9]%', DATA) > 0
AND DATA LIKE '%-%'
AND DATA LIKE '%,%'
Demo on SQLFiddle
As already mentioned u can use a LIKE expression but it will only work with some minor modifications, otherwise too many rows will be filtered out.
SELECT * FROM X WHERE T NOT LIKE '%[^0-9!-,]%' ESCAPE '!'
see working example here:
http://sqlfiddle.com/#!3/474f5/6
edit:
to meet all 3 conditions:
SELECT *
FROM X
WHERE T LIKE '%[0-9]%'
AND T LIKE '%-%'
AND T LIKE '%,%'
see: http://sqlfiddle.com/#!3/86328/1
Maybe not the most beautiful but a working solution.
Related
I have a column in which data has letters with numbers.
For example:
1 name
2 names ....
100 names
When sorting this data, it is not sorted correctly, how can I fix this? I made a request but it doesn’t sort correctly.
select name_subagent
from Subagent
order by
case IsNumeric(name_subagent)
when 1 then Replicate('0', 100 - Len(name_subagent)) + name_subagent
else name_subagent
end
This should work
select name_subagent
from Subagent
order by CAST(LEFT(name_subagent, PATINDEX('%[^0-9]%', name_subagent + 'a') - 1) as int)
This expression will find the first occurrence of a letter withing a string and assume anything prior to this position is a number.
You will need to adapt this statement to your needs as apparently your data is not in Latin characters.
With a bit of tweaking you should be able to achieve exactly what you're looking for:
select
name_subagent
from
Subagent
order by
CAST(SUBSTRING(name_subagent,0,PATINDEX('%[A-Z]%',name_subagent)) as numeric)
Note, the '%[A-Z]%' expression. This will only look for the first occurrence of a letter within the string.
I'm not considering special characters such as '!', '#' and so on. This is the bit you might want to play around with and adapt to your needs.
I have a requirement where I have to find number of records in a special pattern in the field ref_id in a table. It's a varchar column. I need to find all the records where 8th, 9th and 10th character are numeric+XX. That is it should be like 2XX or 8XX. I tried using regexp :digit: but no luck. Essentially I am looking for all records where 8th-10th characters are 1XX, 2XX, 3XX… etc
Using REGEXP_LIKE, replace table with Yours:
SELECT COUNT(*)
FROM table
WHERE REGEXP_LIKE(ref_id,'^.{7}[0-9]XX');
.{7} whatever seven characters
[0-9] 8th character digit
XX 9th and 10th characters X
Or with [:digit:] class as You are mentioning, You may use:
SELECT COUNT(*)
FROM table
WHERE REGEXP_LIKE(ref_id,'^.{7}[[:digit:]]XX');
This can also be achieved using standard non-regex SQL functions
select * from t where s like '________XX%' -- any 8 characters and then XX
AND translate( substr(s,8,1),'?0123456789','?') is null; --8th one is numeric
DEMO
No need for a regexp:
select * from mytable where substr(ref_id, 8, 3) in ('0XX','1XX','2XX','3XX','4XX','5XX','6XX','7XX','8XX','9XX')
or
select * from mytable where substr(ref_id, 8, 3) in ('1XX','2XX','3XX','4XX','5XX','6XX','7XX','8XX','9XX')
I don't know if '0XX' is a valid match or not.
Regexp's tend to be slow.
I have imported a table containing roughly 100,000 records, some of which need to be removed. I'd like to identify and remove any and all records where a particular field (called MyQuery) contains words surrounded in quotation marks, but if there are only TWO quotation marks in the field.
For example I would like to remove
"This is a test"
--but not--
"This is "a" test"
Thank you kindly for any assistance
My apologies to all of you for the terrible structure and clarity of my question. I think I was able to get to the answer based on your suggestions though!
select * from <table> where LEN(myquery) - LEN(REPLACE(myquery,'"','')) = 2 and myquery like '"%"'
Thank you all so much - you're the best!
You can do it like this:
DELETE
-- SELECT * -- To test first!
FROM YourTable
WHERE LEN(MyQuery) - LEN(REPLACE(MyQuery, '"', '')) = 2;
AND MyQuery LIKE '"%"'
You are basically comparing the length if the field after removing all quotes. This is an easy way to determine the number of occurrences of a specific character.
After your comment, I added another the condition so the quotes are always surrounding the field.
You can use the following query to delete all records that contain '"', excluding those that start and end with '"':
DELETE From <TABLENAME> Where COL like '%"%' and (COL not like '"%' Or COL not like '%"')
Or, you can use: Here we delete all records having more than 2 occurrences of "
DELETE FROM <TABLENAME> WHERE (LEN(COL)-len(replace(COL,'"',''))) > 2
Assuming I have table that looks like this:
Id | Name | Age
=====================
1 | Jose | 19
2 | Yolly | 26
20 | Abby | 3
29 | Tara | 4
And my query statement is:
1) Select * from thisTable where Name <= '*Abby';
it returns 0 row
2) Select * from thisTable where Name <= 'Abby';
returns row with Abby
3) Select * from thisTable where Name >= 'Abby';
returns all rows // row 1-4
4) Select * from thisTable where Name >= '*Abby';
returns all rows; // row 1-4
5) Select * from thisTable where Name >= '*Abby' and Name <= "*Abby";
returns 0 row.
6) Select * from thisTable where Name >= 'Abby' and Name <= 'Abby';
returns row with Abby;
My question: why I got these results? How does the wildcard affect the result of query? Why don't I get any result if the condition is this Name <= '*Abby' ?
Wildcards are only interpreted when you use LIKE opterator.
So when you are trying to compare against the string, it will be treated literally. So in your comparisons lexicographical order is used.
1) There are no letters before *, so you don't have any rows returned.
2) A is first letter in alphabet, so rest of names are bigger then Abby, only Abby is equal to itself.
3) Opposite of 2)
4) See 1)
5) See 1)
6) This condition is equivalent to Name = 'Abby'.
When working with strings in SQL Server, ordering is done at each letter, and the order those letters are sorted in depends on the collation. For some characters, the sorting method is much easier to understand, It's alphabetical or numerical order: For example 'a' < 'b' and '4' > '2'. Depending on the collation this might be done by letter and then case ('AaBbCc....') or might be Case then letter ('ABC...Zabc').
Let's take a string like 'Abby', this would be sorted in the order of the letters A, b, b, y (the order they would appear would be according to your collation, and i don't know what it is, but I'm going to assume a 'AaBbCc....' collation, as they are more common). Any string starting with something like 'Aba' would have a value sell than 'Abby', as the third character (the first that differs) has a "lower value". As would a value like 'Abbie' ('i' has a lower value than 'y'). Similarly, a string like 'Abc' would have a greater value, as 'c' has a higher value than 'b' (which is the first character that differs).
If we throw numbers into the mix, then you might be surpised. For example the string (important, I didn't state number) '123456789' has a lower value than the string '9'. This is because the first character than differs if the first character. '9' is greater than '1' and so '9' has the "higher" value. This is one reason why it's so important to ensure you store numbers as numerical datatypes, as the behaviour is unlikely to be what you expect/want otherwise.
To what you are asking, however, the wildcard for SQL Server is '%' and '_' (there is also '^',m but I won't cover that here). A '%' represents multiple characters, while '_' a single character. If you want to specifically look for one of those character you have to quote them in brackets ([]).
Using the equals (=) operator won't parse wildcards. you need to use a function that does, like LIKE. Thus, if you want a word that started with 'A' you would use the expression WHERE ColumnName LIKE 'A%'. If you wanted to search for one that consisted of 6 characters and ended with 'ed' you would use WHERE ColumnName LIKE '____ed'.
Like I said before, if you want to search for one of those specific character, you quote then. So, if you wanted to search for a string that contained an underscore, the syntax would be WHERE ColumnName LIKE '%[_]%'
Edit: it's also worth noting that, when using things like LIKE that they are effected by the collations sensitivity; for example, Case and Accent. If you're using a case sensitive collation, for example, then the statement WHERE 'Abby' LIKE 'abb%' is not true, and 'A' and 'a' are not the same case. Like wise, the statement WHERE 'Covea' = 'Covéa' would be false in an accent sensitive collation ('e' and 'é' are not treated as the same character).
A wildcard character is used to substitute any other characters in a string. They are used in conjunction with the SQL LIKE operator in the WHERE clause. For example.
Select * from thisTable WHERE name LIKE '%Abby%'
This will return any values with Abby anywhere within the string.
Have a look at this link for an explanation of all wildcards https://www.w3schools.com/sql/sql_wildcards.asp
It is because, >= and <= are comparison operators. They compare string on the basis of their ASCII values.
Since ASCII value of * is 42 and ASCII values of capital letters start from 65, that is why when you tried name<='*Abby', sql-server picked the ASCII value of first character in your string (that is 42), since no value in your data has first character with ASCII value less than 42, no data got selected.
You can refer ASCII table for more understanding:
http://www.asciitable.com/
There are a few answers, and a few comments - I'll try to summarize.
Firstly, the wildcard in SQL is %, not * (for multiple matches). So your queries including an * ask for a comparison with that literal string.
Secondly, comparing strings with greater/less than operators probably does not do what you want - it uses the collation order to see which other strings are "earlier" or "later" in the ordering sequence. Collation order is a moderately complex concept, and varies between machine installations.
The SQL operator for string pattern matching is LIKE.
I'm not sure I understand your intent with the >= or <= stateements - do you mean that you want to return rows where the name's first letter is after 'A' in the alphabet?
How can i query a column with Names of people to get only the names those contain exactly 2 “a” ?
I am familiar with % symbol that's used with LIKE but that finds all names even with 1 a , when i write %a , but i need to find only those have exactly 2 characters.
Please explain - Thanks in advance
Table Name: "People"
Column Names: "Names, Age, Gender"
Assuming you're asking for two a characters search for a string with two a's but not with three.
select *
from people
where names like '%a%a%'
and name not like '%a%a%a%'
Use '_a'. '_' is a single character wildcard where '%' matches 0 or more characters.
If you need more advanced matches, use regular expressions, using REGEXP_LIKE. See Using Regular Expressions With Oracle Database.
And of course you can use other tricks as well. For instance, you can compare the length of the string with the length of the same string but with 'a's removed from it. If the difference is 2 then the string contained two 'a's. But as you can see things get ugly real soon, since length returns 'null' when a string is empty, so you have to make an exception for that, if you want to check for names that are exactly 'aa'.
select * from People
where
length(Names) - 2 = nvl(length(replace(Names, 'a', '')), 0)
Another solution is to replace everything that is not an a with nothing and check if the resulting String is exactly two characters long:
select names
from people
where length(regexp_replace(names, '[^a]', '')) = 2;
This can also be extended to deal with uppercase As:
select names
from people
where length(regexp_replace(names, '[^aA]', '')) = 2;
SQLFiddle example: http://sqlfiddle.com/#!4/09bc6
select * from People where names like '__'; also ll work